\chapter{Exponentials and logarithms}
\section{Graphs of exponential and logarithmic functions}
In Unit 3.1 we drew the graph of the exponential function \(y=b^x\).
If the base \(b\) is greater than \(1\), the graph is always increasing.
If the base is
less than \(1\) (that is often written by making the exponent negative),
the graph is always decreasing.
The two most important examples of each case are the exponential functions
\(y=e^x\) and \(y=e^{-x}\), whose graphs are shown below.
\[ \img{U3_1F4.png}{}{12em}{} \hspace{5ex} \img{U3_6F1.png}{}{12em}{} \]
We also listed their properties:
| Properties of exponential functions \(y=b^x\) |
|---|
|
\begin{itemize}
\item Domain \(=(-\infty,\infty)\)
\item \(x\)-intercept: None
\item Vertical asymptote: None
\item One-to-one: Yes
\end{itemize} |
\begin{itemize}
\item Range \(=(0,\infty)\)
\item \(y\)-intercept: \((0,1)\)
\item Horizontal asymptote: \(y=0\) (the \(x\)-axis)
\item Increasing if \(b> 1\), decreasing if \(b < 1\).
\end{itemize} |
|
In this unit we will consider these functions as new parent functions and apply
the usual simple transformations (shifting, stretching or compressing, and reflecting) to get many more functions,
and we will draw the graph of the transformed functions.
Since the domain is \((-\infty, \infty)\), all transformed functions will also have the same domain.
But range, intercepts and asymptotes will change as we transform the
parent function. But the basic shape of the function (always increasing or always decreasing,
and one-to-one) will remain the same.
\subsection{Transformed exponential functions}
\begin{example}
We are given the transformed exponential function
\[f(x)=2e^{x-1} -3.\]
This function is obtained from the parent function \(y=e^x\) by stretching vertically by
\(2\), shifting right by \(1\) and down by \(3\).
As we found in Unit 1.11, it is useful to identify a base point \(P\) that will help us keep track of
the transformation. We use the \(y\)-intercept \((0,1)\) of the parent function
as the base point \(P\), and we denote by \(P'\) the transformed point.
The following pictures show the graph of the parent \(y=e^x\)
as a dashed line, and the transformation of the base point \(P\) and the horizontal asymptote.
\textbf{Important:} Vertical shifts of these graphs will move the horizontal asymptote in the same way that they move the graphs of the functions. Remember to shift the asymptote along with the graphs of the functions!
\[
\begin{array}{cc}
\img{U3_6F2.png}{}{12em}{} \hspace{1ex} & \longrightarrow \hspace{1ex}
\img{U3_6F3.png}{}{12em}{}\\
& \mbox{stretch vertically by \(2\)}\\[2ex]
\longrightarrow \hspace{1ex} \img{U3_6F4.png}{}{12em}{} &\\
\hspace{3ex}\mbox{shift right by \(1\) and down by \(3\)} &
\end{array}\]
Note that the horizontal asymptote \(y=0\) gets shifted down to \(y=-3\).
It is clear from the position of the transformed point \(P'\) that the graph will have an \(x\)-
intercept, as well as a different \(y\)-intercept than the parent function. So before drawing
the transformed graph, we find the intercepts.
For the \(y\)-intercept, we compute \(f(0)\):
\[f(0)=2e^{0-1}-3=2e^{-1}-3=\frac{2}{e}-3\]
For the \(x\)-intercept, we need to solve the equation \(f(x)=0\):
\[\begin{array}{rcll}
2e^{x-1}-3 & = & 0 \\[1ex]
2e^{x-1} &=& 3 & \mbox{Add \(3\) to both sides}\\[1ex]
e^{x-1} &=& \displaystyle \frac{3}{2} & \mbox{Divide by \(2\)}\\[1ex]
x-1 &=& \displaystyle \ln \left(\frac{3}{2}\right) & \mbox{Use the translation for \(\ln\)}\\[2ex]
x &=& 1+ \ln (3/2) &\mbox{Add \(1\) to both sides}
\end{array}\]
So we found:
\[\mbox{\(y\)-intercept: } \left(0,\frac{2}{e}-3\right) , \hspace{5ex}
\mbox{\(x\)-intercept: } (1+\ln(3/2),0).\]
These are the exact values of the intercepts, and they are irrational numbers. In order
to use them for the graph we will need to find a decimal approximation with a calculator:
\[\mbox{\(y\)-intercept: } \left(0,\frac{2}{e}-3\right) \approx (0,-2.3), \hspace{5ex}
\mbox{\(x\)-intercept: } (1+\ln(3/2),0) \approx (1.4,0).\]
To draw a more accurate graph, we also find a few points using a calculator:
\[
\begin{array}{c|ccccc}
x & -2 & -1 & 1 & 2 \\
\hline
f(x) & -2.9 & -2.7 & -1 & 2.4
\end{array}
\]
We can now draw the graph, showing the intercepts and the asymptote:
\[\img{U3_6F5.png}{}{12em}{}\]
We see from the graph that the range of \(f(x)\) is \(R=(-3,\infty)\), while the domain
is the same as that of the parent function, \(D=(-\infty,\infty)\).
\end{example}
\subsection{Transformed logarithmic functions}
We now use the logarithmic function \(y=\log_b x\) as parent function and discuss its
transformations. We will only treat the case \(b > 1\). The most common case is \(y=\ln x\),
the natural logarithm in base \(e\). The graph and properties (as done in Unit 3.3) are shown
below.
\[ \img{U3_6F15.png}{}{12em}{} \]
| Properties of logarithmic functions \(y=\log_b x\), \(b\gt 1\) |
|---|
|
\begin{itemize}
\item Domain \(=(0,\infty)\)
\item \(x\)-intercept: \((1,0)\)
\item Vertical asymptote: \(x=0\) (the \(y\)-axis)
\item One-to-one: Yes
\end{itemize} |
\begin{itemize}
\item Range \(=(-\infty,\infty)\)
\item \(y\)-intercept: None
\item Horizontal asymptote: None
\item Increasing
\end{itemize} |
|
Remember that the properties of \(y=\ln x\) are obtained from those of \(y=e^x\) by interchanging
\(x\) and \(y\), so that the \(y\)-intercept \((0,1)\) becomes the \(x\)-intercept \((1,0)\),
the domain becomes the range, etc.
\begin{example}
We are given the function \(f(x)=\log_3(x+1)-2\). We will draw the graph and derive the
properties. The parent function is \(y=\log_3x\), then there is a shift to the left by \(1\)
and down by \(2\).
We will use the \(x\)-intercept \((1,0)\) as base point \(P\) to keep track of
the transformation, and we denote by \(P'\) the transformed point.
The following pictures show the graph of the parent \(y=\log_3 x\)
as a dashed line, and the transformation of the base point \(P=(1,0)\) and the vertical
asymptote \(x=0\).
\textbf{Important:} Horizontal shifts of these graphs will move the vertical asymptote in the same way that they move the graph
of the functions. Remember to shift the asymptote along with the graphs of the functions!
\[
\begin{array}{ccc}
\img{U3_6F16.png}{}{12em}{} \hspace{1ex} & \longrightarrow \hspace{1ex} &
\img{U3_6F17.png}{}{12em}{}\\
& & \mbox{shift left by \(1\) and down by \(2\)}
\end{array}\]
Note that the vertical asymptote \(x=0\) gets shifted left to \(x=-1\).
It is clear from the position of the transformed point \(P'\) that the graph will have a \(y\)-
intercept, as well as a different \(x\)-intercept than the parent function. So before drawing
the transformed graph, we find the intercepts.
To find the \(y\)-intercept, we compute
\[f(0)=\log_3(0+1) -2=\log_3 1 -2= 0-2=-2.\]
So the \(y\)-intercept is \((0,-2)\). For the \(x\)-intercept, we solve \(f(x)=0\):
\[\begin{array}{rcll}
\log_3(x+1)-2 & = & 0 \\[2ex]
\log_3(x+1) &=& 2 & \mbox{Add \(2\) to both sides}\\[2ex]
x+1 &=& 3^2 & \mbox{Use the translation formula}\\[2ex]
x &=& 9-1 & \mbox{Simplify and subtract \(1\) from both sides}\\[2ex]
x &=& 8 &
\end{array}\]
So the \(x\)-intercept is \((8,0)\).
We find a few points with a calculator, and draw the graph:
\[
\begin{array}{c|ccccc}
x & 1 & 2 & 3 & 4 \\
\hline
f(x) & -1.4 & -1 & -0.7 & -0.5
\end{array}
\]
\[\img{U3_6F18.png}{}{12em}{}\]
The domain is \(D=(-1,\infty)\) and the range is \(R=(-\infty,\infty)\).
\end{example}
Problems
\problem
Let \(f(x)=-2^{x+2}+3\).
\begin{enumerate}
\item
Identify the parent function, and the transformations used to get \(f(x)\).
\item
Find all intercepts of \(f(x)\), both exact values and a decimal approximation.
\item
Find all asymptotes of \(f(x)\).
\item
Draw an accurate graph of \(f(x)\), using a few extra points and showing all intercepts and asymptotes.
\item
Find domain and range of \(f(x)\).
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
The parent function is \(y=2^x\). The transformations are:
\begin{itemize}
\item
Reflect across the \(x\)-axis
\item
Shift left by \(2\) and up by \(3\).
\end{itemize}
\item
For the \(y\)-intercept, we compute \(f(0)\):
\[f(0)=-2^{0+2}+3=-2^2+3 =-4+3=-1.\]
So the \(y\) intercept is \((0,-1)\).
For the \(x\)-intercept, we solve \(f(x)=0\):
\[
\begin{array}{rcll}
-2^{x+2}+3&=& 0 \\[2ex]
3&=& 2^{x+2} & \mbox{add \(2^{x+2}\) to both sides}\\[2ex]
\ln 3 &=& \ln 2^{x+2} & \mbox{Take \(\ln\) of both sides}\\[2ex]
\ln 3 &=& (x+2)\ln 2 &\mbox{Use property (C) of logarithms}\\[2ex]
\displaystyle \frac{\ln 3}{\ln 2}&=& x+2 & \mbox{Divide both sides by \(\ln 2\), and simplify}\\[2ex]
\displaystyle \frac{\ln 3}{\ln 2} -2 & = & x & \mbox{Subtract \(2\) from both sides}.
\end{array}
\]
So the \(x\)-intercept is \(\displaystyle \left(\frac{\ln 3}{\ln 2} -2,0\right)\approx -0.4\).
\item
Exponential functions and their transformations have no vertical asymptotes. The horizontal
asymptote is found by shifting the horizontal asymptote \(y=0\) of the parent function
up by \(3\). So the horizontal asymptote is \(y=3\).
\item
We can now draw the graph. First we reflect \(y=2^x\) across the \(x\)-axis, so the base point
\(P\) is moved to \((0,-1)\):
\[\img{U3_6F7.png}{}{12em}{}\]
Then we track the location of the transformed base point \(P'\) and the horizontal asymptote, shifting
left \(2\) and up \(3\):
\[\img{U3_6F8.png}{}{12em}{}\]
Finally we find a few extra points, and draw the graph:
\[
\begin{array}{c|cccc}
x & -3 & -2 & -1 & 1 \\
\hline
f(x) & 2.5 & 2 & 1 & -5
\end{array}
\]
\[\img{U3_6F6.png}{}{12em}{}\]
\item
We see from the graph that the range is \(R=(-\infty,3)\), and the domain is \((-\infty,\infty)\).
\end{enumerate}
\end{sol}
\mproblem
Let \(f(x)=-e^{x+1}+2\).
\begin{enumerate}
\item
Identify the parent function, and the transformations used to get \(f(x)\).
\item
Find all intercepts of \(f(x)\), both exact values and a decimal approximation.
\item
Find all asymptotes of \(f(x)\).
\item
Draw an accurate graph of \(f(x)\), using a few extra points and showing all intercepts and asymptotes.
\item
Find domain and range of \(f(x)\).
\end{enumerate}
\problem
Let \(f(x)=2e^{-x}+1\).
\begin{enumerate}
\item
Identify the parent function, and the transformations used to get \(f(x)\).
\item
Find all intercepts of \(f(x)\), both exact values and a decimal approximation.
\item
Find all asymptotes of \(f(x)\).
\item
Draw an accurate graph of \(f(x)\), using a few extra points and showing all intercepts and asymptotes.
\item
Find domain and range of \(f(x)\).
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
The parent function is the decreasing exponential \(y=e^{-x}\). The transformations are: stretch
vertically by \(2\), and shift up by \(1\).
\item
For the \(y\)-intercept, we find
\[f(0)=2e^{-0}+1=2(1)+1=3.\]
For the \(x\)-intercept, we solve \(f(x)=0\):
\[\begin{array}{rcll}
2e^{-x}+1&=& 0\\[2ex]
2e^{-x} &=& -1 &\mbox{Subtract \(1\) from both sides}\\[2ex]
e^{-x} & = & \displaystyle \frac{-1}{2} & \mbox{Divide both sides by \(2\)}\\
& & & \mbox{An exponential cannot be negative}\\[2ex]
& ?? & & \mbox{No solutions}
\end{array}
\]
So the \(y\)-intercept is \((0,3)\), and there are no \(x\)-intercepts.
\item
There are no vertical asymptotes, and shifting up by \(1\) the horizontal asymptote \(y=0\)
of the parent function, we find that the horizontal asymptote of \(f(x)\) is \(y=1\).
\item
To draw the graph, we first stretch the base point \(P=(0,1)\) by \(2\), then we shift up
by \(1\):
\[
\begin{array}{cc}
\img{U3_6F9.png}{}{12em}{} \hspace{1ex} & \longrightarrow \hspace{1ex}
\img{U3_6F10.png}{}{12em}{}\\
& \mbox{stretch vertically by \(2\)}\\[2ex]
\longrightarrow \hspace{1ex} \img{U3_6F11.png}{}{12em}{} &\\
\hspace{3ex}\mbox{shift up by \(1\)} &
\end{array}\]
We find a few points with the calculator:
\[
\begin{array}{c|ccccc}
x & -1 & 1 & 2 & 3 \\
\hline
f(x) & 6.4 & 1.7 & 1.3 & 1.1
\end{array}
\]
We draw the graph:
\[\img{U3_6F12.png}{}{12em}{}\]
\item
The domain is \(D=(-\infty,\infty)\) and the range is \(R=(1,\infty)\).
\end{enumerate}
\end{sol}
\mproblem
Let \(f(x)=2^{-x}-3\).
\begin{enumerate}
\item
Identify the parent function, and the transformations used to get \(f(x)\).
\item
Find all intercepts of \(f(x)\), both exact values and a decimal approximation.
\item
Find all asymptotes of \(f(x)\).
\item
Draw an accurate graph of \(f(x)\), using a few extra points and showing all intercepts and asymptotes.
\item
Find domain and range of \(f(x)\).
\end{enumerate}
\problem
Let \(f(x)=2\ln(x-2)+3\).
\begin{enumerate}
\item
Identify the parent function, and the transformations used to get \(f(x)\).
\item
Find all intercepts of \(f(x)\), both exact values and a decimal approximation.
\item
Find all asymptotes of \(f(x)\).
\item
Draw an accurate graph of \(f(x)\), using a few extra points and showing all intercepts and asymptotes.
\item
Find domain and range of \(f(x)\).
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
The parent function is \(y=\ln x\). The transformations are:
\begin{itemize}
\item
Stretch vertically by \(2\)
\item
Shift right by \(2\) and up by \(3\).
\end{itemize}
\item
The parent function \(\ln x\) has the \(y\)-axis as vertical asymptote, and \(f(x)\) is shifted
to the right by \(2\). So it cannot have any \(y\)-intercept. We can also see this from the
fact that if we try to compute \(f(0)\), we find \(\ln(0-2)=\ln(-2)\), that is undefined.
For the \(x\)-intercept, we solve \(f(x)=0\):
\[
\begin{array}{rcll}
2\ln(x-2)+3&=& 0 \\[2ex]
2\ln(x-2)&=& -3 & \mbox{Subtrat \(3\) from both sides}\\[2ex]
\ln (x-2) &=& -3/2 & \mbox{Divide both sides by \(2\)}\\[2ex]
x-2 &=& e^{-3/2} &\mbox{Use the translation formula}\\[2ex]
x&=& e^{-3/2} +2 & \mbox{Add \(2\) to both sides}\\[2ex]
x & = & \displaystyle \frac{1}{\sqrt{e^3}}+2 & \mbox{Re-write as radical}\\[2ex]
x&=& \displaystyle \frac{1}{e\sqrt{e}}+2& \mbox{Simplify}.
\end{array}
\]
So the \(x\)-intercept is \(\displaystyle \left(\frac{1}{e\sqrt{e}} +2,0\right)\approx 2.2\).
\item
Logarithmic functions and their transformations have no horizontal asymptotes. The vertical
asymptote is found by shifting the vertical asymptote \(x=0\) of the parent function
right by \(2\). So the vertical asymptote is \(x=2\).
\item
We can now draw the graph. Stretching does not change the base point, because its \(y\)-coordinate
is \(0\), and \(2(0)=0\). Shifting right by \(2\) and up by \(3\) we find the transformed point
\(P'=(3,3)\) and the shifted vertical asymptote \(x=2\). The following picture shows
\(P'\), the shifted vertical asymptote, and the \(x\)-intercept:
\[\img{U3_6F19.png}{}{12em}{}\]
We find a few points with a calculator, and draw the graph:
\[
\begin{array}{c|ccccccc}
x & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
f(x) & 4.4 & 5.2 & 5.8 & 6.2 & 6.6 & 6.9
\end{array}
\]
\[\img{U3_6F20.png}{}{12em}{}\]
\item The domain is \(D=(2,\infty)\), and the range is \(R=(-\infty,\infty)\).
\end{enumerate}
\end{sol}
\mproblem
Let \(f(x)=-\log_2(x+2)+1\).
\begin{enumerate}
\item
Identify the parent function, and the transformations used to get \(f(x)\).
\item
Find all intercepts of \(f(x)\), both exact values and a decimal approximation.
\item
Find all asymptotes of \(f(x)\).
\item
Draw an accurate graph of \(f(x)\), using a few extra points and showing all intercepts and asymptotes.
\item
Find domain and range of \(f(x)\).
\end{enumerate}